Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-6x+3y &= -6 \\ 9x-9y &= 3\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $9x = 9y+3$ Divide both sides by $9$ to isolate $x$ $x = {y + \dfrac{1}{3}}$ Substitute this expression for $x$ in the first equation. $-6({y + \dfrac{1}{3}}) + 3y = -6$ $-6y - 2 + 3y = -6$ Simplify by combining terms, then solve for $y$ $-3y - 2 = -6$ $-3y = -4$ $y = \dfrac{4}{3}$ Substitute $\dfrac{4}{3}$ for $y$ in the top equation. $-6x+3( \dfrac{4}{3}) = -6$ $-6x+4 = -6$ $-6x = -10$ $x = \dfrac{5}{3}$ The solution is $\enspace x = \dfrac{5}{3}, \enspace y = \dfrac{4}{3}$.